If A, B and C are the interior angles of a triangle ABC, show that sin\(B+C\over 2\) = cos\(A\over 2\).

Solution :

Since A, B and C are the interior angles of a triangle ABC

\(\therefore\)   A + B + C = 180

\(\implies\)  \(A\over 2\) + \(B\over 2\) + \(C\over 2\) = 90

\(\implies\)  \(B\over 2\) + \(C\over 2\) = 90 – \(A\over 2\)

\(\implies\)  sin(\(B+C\over 2\)) = sin(90 – \(A\over 2\))

\(\implies\)   sin\(B+C\over 2\) = cos\(A\over 2\).

Leave a Comment

Your email address will not be published.